EQUATIONS INVOLVING ARITHMETIC FUNCTIONS OF FIBONACCI AND LUCAS NUMBERS Florian Luca Mathematical Institute, Czech Academy of Sciences Zitna 25, 115 67 Praha 1, Czech Republic (Submitted March 1998-Final Revision November 1998) For any positive integer k, let <f>(k) and a(k) be the number of positive integers less than or equal to k and relatively prime to k and the sum of divisors ofk, respectively. In [6] we have shown that 0(Fn) > F^n) and that a(Fn) < Fa{ri) and we have also determined all the cases in which the above inequalities become equalities. A more general inequality of this type was proved in [7]. In [8] we have determined all the positive solutions of the equation $(xm - ym) = x" + y" and in [9] we have determined all the integer solutions of the equation $(\xm + ym |) = \xn + yn\. In this paper, we present the following theorem. Theorem: (1) The only solutions of the equation m H\Fn\) = 2 , (1) are obtained for n = +1, ± 2, ± 3, + 4, ± 5, ± 6, ± 9. (2) The only solutions of the equation *(l4l) = 2m, (2) are obtained for n = 0, ± 1, ± 2, ± 3. (3) The only solutions of the equation ^(|FJ) = 2"-, (3) are obtained for n = ±1, ± 2, ± 4, ± 8. (4) The only solutions of the equation ^(I4l) = 2m, (4) are obtained for n = ±1, ± 2, ± 4. Let n > 3 be a positive integer. It is well known that the regular polygon with n sides can be constructed with the ruler and the compass if and only if $(n) is a power of 2. Hence, the above theorem has the following immediate corollary. Comllaiy: (1) The only regular polygons that can be constructed with the ruler and the compass and whose number of sides is a Fibonacci number are the ones with 3, 5, 8, and 34 sides, respectively. (2) The only regular polygons that can be constructed with the ruler and the compass and whose number of sides is a Lucas number are the ones with 3 and 4 sides, respectively. The question of finding all the regular polygons that can be constructed with the ruler and the compass and whose number of sides n has various special forms has been considered by us 2000] 49 EQUATIONS INVOLVING ARITHMETIC FUNCTIONS OF FIBONACCI AND LUCAS NUMBERS previously. For example, in [10] we found all such regular polygons whose number of sides n belongs to the Pascal triangle and in [11] we found all such regular polygons whose number of sides n is a difference of two equal powers. We begin with the following lemmas. Lemma 1: l (1) F_n = (-\r Fn and L_n = (-1)"4. 2 2F ( ) m+n = FmK + LnFm and 2Lm+n = 5FmF„ + LmLn. l (3) F2n=FnLnmdL2n = 4 + 2(-ir . 2 2 (4) L n-5Fn = 4(-iy. Proof: See [2]. • Lemma 2: (1) Let p > 5 be a prime number. If (7) = 1, then p \ Fp_x. Otherwise, p \ Fp+l. (2) (Fm, Fn) = i ^ W) for all positive integers m and n. (3) If rn \n and nlm is odd, then Lm \Ln. (4) Let p and n be positive integers such that p is an odd prime. Then ( i Fn)>^ if and only if /? I w and nl pis even. Proof: (1) follows from Theorem XXII in [1]. (2) follows either from Theorem VI in [1] or from Theorem 2.5 in [3] or from the Main Theorem in [12]. (3) follows either from Theorem VII in [1] or from Theorem 2.7 in [3] or from the Main Theorem in [12]. (4) follows either from Theorem 2.9 in [3] or from the Main Theorem in [12]. • Lemma 3: Let k > 3 be an integer. k k 1 (1) The period of {Fn)n>o modulo 2 is 2 ' -3. k k+l k k 2 (2) F2k_2.3 = 2 (mod 2 ). Moreover, if Fn s 0 (mod 2 \ then n = 0 (mod 2 ~ • 3). k k fjj Assume that n is an odd integer such that Fn = ±l (mod 2 ). Then i^ = 1 (mod 2 ) and w = ±l(mod2^1-3). Proof: (1) follows from Theorem 5 in [13]. (2) The first congruence is Lemma 1 in [4]. The second assertion follows from Lemma 2 in [5]. k (3) We first show that Fn£-\ (mod 2 ). Indeed, by (1) above and the Main Theorem in [4], k X: 1 it follows that the congruence Fn = -1 (mod 2 ) has only one solution n (mod2 ~ -3). Since k l F_2 = - 1 , it follows that n = -2 (mod 2 ~ • 3). This contradicts the fact that n is odd. k We now look at the congruence Fn = 1 (mod 2 ). By (1) above and the Main Theorem in [4], -1 it follows that this congruence has exactly three solutions n (mod2* -3). Since F_l=Fl = F2 = l, it follows that n = ±1,2 (mod 2k~l • 3). Since n is odd, it follows that n = ±1 (mod 2k~l -3). • 50 [FEB. EQUATIONS INVOLVING ARITHMETIC FUNCTIONS OF FIBONACCI AND LUCAS NUMBERS Lemma 4: Let k > 3 be a positive integer. Then [2*+13-l(mod2*+4) if k = 1 (mod 2), 2* " [2*+15-1 (mod2k+4) ifk^O(mod2). Proof: One can check that the asserted congruences hold for k = 3 and 4. We proceed by induction on k. Assume that the asserted congruence holds for some k > 3. +1 k+4 Suppose that k is odd. Then L2k = 2* 3 - 1 + 2 l for some integer /. Using Lemma 1(3), it follows that fc+1 2 +5 +1 2 2 +8 2 L2k+l = L\k - 2 = ((2 3 -1) + 2* /(2* 3 -1) + 2 * / ) - 2 Es(2*+13-l)2-2(mod2*+5). Hence, 2 +2 +2 2 + 2 +2 k+5 L2k+l SE2 * 9-2* 3 + 1-2E2 * 9 + 2* (-3)-1 (mod2 ). Since k>3. it follows that 2k + 2 > k + 5. Moreover, since -3 = 5(mod23), the above congru- ence becomes fc+2 5 L2&+1=2 5-l(mod2^ ). The case k even can be dealt with similarly. D Proof of the Theorem: In what follows, we will always assume that n > 0. m (1) We first show that if $(Fn) = 2 , then the only prime divisors of n are among the elements of the set {2,3,5}. Indeed, assume that this is not the case. Let p> 5 be a prime m m number dividing n. Since Fp\Fm it follows that <f>(Fp)\</>(Fn) = 2 . Hence, <f>{Fp) = 2 K It follows that l Fp = 2 Pl A , (5) where / > 0, k > 0, and pi < p2 < • *• < pk are Fermat primes. Notice that / = 0 and pl > 5. Indeed, since p > 5 is a prime, it follows, by Lemma 2(2), that Fp is coprime to Fm for l<m<5. Since F3 = 2,F4 = 3, and F5 = 5, it follows that 1 = 0 and Pl>5. 2 Hence, pi > 5 for all i = 1,..., *. Write pf = 2 "' + for some at > 2. It follows that 2a,-1 A = 4 +l = 2(mod5). Since (%f) = (f) = - 1 , it follows, by the quadratic reciprocity law, that (-^) = - 1 . It follows, by Lemma 2(1), that px \Fpi+l. Hence, Pi\(Fp,Fpi+l) = FiPfPi+iy 2a,_1 The above divisibility relation and the fact that p is prime, forces p \ px +1 = 2(2 +1). Hence, 2 a , 1 / ? |2 - + l. Thus, p<22°x'l^l (6) On the other hand, since ^ = fl(22a' +D-l(mod22a'), 2000] 51 EQUATIONS INVOLVING ARITHMETIC FUNCTIONS OF FIBONACCI AND LUCAS NUMBERS it follows, by Lemma 3(3), that p = ±1 (mod 22"1 *3). In particular, p>22a'-l3-l (7) From inequalities (6) and (7), it follows that 22*'~13 - 1 < 22"'"1 +1 or 22"1 < 2. This implies that ax - 0 which contradicts the fact that al>2. a b c Now write n = 2 3 S . We show that a < 2. Indeed, if a > 3, then 21 = Fs \ Fn, therefore 3|12=0(21)|0(FJ = 2'», which is a contradiction. We show that b < 2. Indeed, if b > 3, then 531F27 | i^, therefore 13|52 = *(53)|rt/y = 2», which is a contradiction. Finally, we show that c < 1. Indeed, if c > 2, then 30011F251 i^, there- fore 313000 = 0(3001) |0(FJ = 2W, which is again a contradiction. In conclusion, w |22 -32 -5 = 180. One may easily check that the only divisors n of 180 for which (f>{Fn) is a power of 2 are indeed the announced ones. (2) Since 0(2) = 0(1) = 1 = 2° and 0(3) = 0(4) = 21, it follows that n = 0,1,2,3 lead to solutions of equation (2). We now show that these are the only ones. One may easily check that W n * 4,5. Assume that n>6. Since </>(Ln) = 2 , it follows that l Ln = 2 -Pl—pk, (8) r 2 Ci where / > 0 and P\<"'<Pk &e Fermat primes. Write pi=2 ° -hi. Clearly, pt>3. The sequence {Ln)n>0 is periodic modulo 8 with period 12. Moreover, analyzing the terms Ls for 5 = 0,1,..., 11, one notices that Ls^0 (mod 8) for any 5 = 0,1,..., 11. It follows that / < 2 in equation (8). Since /i>6, it follows that Zw>18. In particular, p>5 for some / = 1,...,£. From the equation 2 2 L n-5F„ = (-l)"-4, (9) 2 it follows easily that 5|ZW. Thus, pi > 5. Hence, pi = 2 "' +1 for some af>2.